Question

If the nth term of an AP is (4n + 1), find the sum or the first 15 terms of this AP. Also find the sum of its n terms.

Answer 1

Given: T_n = (4n + 1).
Now, T₁ = (4 ⨯ 1 + 1) = 5
T₂ = (4 ⨯ 2 + 1) = 9
T₁₅ = (4 ⨯ 15 + 1) = 61
Hence, a = 5, d = (9 - 5) = 4 and l = 61

∴ S₁₅ = $\frac{n}{2}\left(a+l\right)$
$=\left(\frac{15}{2}\right)\times \left(5+61\right)=15\times 33=495$
Hence, the sum of the first 15 terms is 495.
The sum of the first n terms is given by
$$\begin{gathered} S_n=\frac{n}{2}\left[2a+\left(n-1\right)d\right] \\ =\frac{n}{2}\left[2\times 5+\left(n-1\right)\times 4\right]=\frac{n}{2}\left[6+4n\right]=\left(3n+2n^2\right) \end{gathered}$$